Video Transcript
A transistor is connected in a circuit such that 𝑉 sub CC equals seven volts, 𝑅 sub
C equals three kiloohms, 𝑉 sub CE equals 0.4 volts, and 𝛽 sub e equals 27. Calculate the base current and give your answer to three decimal places in
milliamperes.
Here, we’ve been given a list of values. So let’s start by recalling what each one represents. It’ll also be helpful to draw a circuit diagram that we can label the values on in
order to help us better understand the quantities that we’re dealing with. Notice that we’ve labeled the three regions of the transistor too: the collector,
base, and emitter.
The first value we’ve been given, 𝑉 sub CC, is the potential difference supplied
across the collector and the emitter by this cell down here. 𝑅 sub C is the resistance to the collector current, and 𝑉 sub CE is the potential
difference measured across the collector and the emitter. Notice that the current going into the collector, which is known as the collector
current, passes through this resistor. This means that the potential difference 𝑉 sub CC that is supplied by the cell gets
shared between the resistor and the transistor. That’s why the value of the potential difference 𝑉 sub CE measured across the
collector and emitter of the transistor is less than the value of 𝑉 sub CC.
Now, the other piece of information that we’re given is the value of 𝛽 sub e, which
represents the current gain of the circuit. Let’s recall that its value can be found by dividing the collector current 𝐼 sub C
by the base current 𝐼 sub B.
Now, this question is asking us to solve for the base current; that’s this current in
our diagram. So let’s rearrange this formula to make 𝐼 sub B the subject. Multiplying both sides by 𝐼 sub B over 𝛽 sub e and canceling the 𝛽 sub e terms on
the left and the 𝐼 sub B terms on the right, we have that the base current 𝐼 sub B
is equal to the collector current 𝐼 sub C divided by the current gain 𝛽 sub e.
So, we have an expression for the quantity that we want, 𝐼 sub B. But we don’t yet know the value of the collector current 𝐼 sub C in this
expression. That means we’ll need to find a way to express 𝐼 sub C in terms of values that we do
know.
To start with, we know that the collector current must pass through the collector
resistor, whose resistance we do know is given by 𝑅 sub C. Then, by Ohm’s law, we know that multiplying 𝐼 sub C by 𝑅 sub C gives the potential
difference across that resistor. Notice that we can relate this value to other known values we’ve been given by
applying Kirchhoff’s second law to the outermost loop of this circuit. Remember that this law states that the sum of the potential difference across each
component in a loop in a circuit is equal to zero.
So starting at this potential difference source and working our way around the
outside loop of the circuit, we find from Kirchhoff’s second law that zero equals 𝑉
sub CC minus 𝐼 sub C times 𝑅 sub C minus 𝑉 sub CE. That is, the potential difference provided by the cell minus the potential difference
across the resistor and minus the potential difference across the transistor is
equal to zero.
We can rearrange this equation to get an expression for the collector current 𝐼 sub
C in terms of the quantities 𝑉 sub CC, 𝑅 sub C, and 𝑉 sub CE whose values we
know. To do this, we’ll first add 𝐼 sub C times 𝑅 sub C to both sides, noting that we can
then cancel the equal positive and negative terms on the right-hand side. Then, we can divide both sides by 𝑅 sub C, which leaves us with an equation that
says the collector current 𝐼 sub C is equal to 𝑉 sub CC minus 𝑉 sub CE divided by
𝑅 sub C.
Remember though that we really want to solve for the base current 𝐼 sub B. We already found this expression for 𝐼 sub B in terms of 𝐼 sub C and 𝛽 sub e. So now, let’s clear some space so that we can substitute in this expression we’ve got
for 𝐼 sub C.
Subbing in our expression for the collector current 𝐼 sub C, we get this expression
for the base current 𝐼 sub B. We can rewrite this more simply as 𝑉 sub CC minus 𝑉 sub CE over 𝑅 sub C times 𝛽
sub e. We know the values of all the quantities on the right-hand side of this
expression. However, before we substitute those values in, we need to convert the resistance 𝑅
sub C from kiloohms into ohms. Recalling that one kiloohm is equal to 1000 ohms, we see that three kiloohms must be
equal to 3000 ohms. Then, substituting in our values, we get this expression here. Typing this into a calculator, we get a result of 8.148 recurring times 10 to the
negative five amperes.
Notice though that we’re asked for our answer in units of milliamperes to three
decimal places. Recall that one ampere is equal to 1000 milliamperes. So in units of milliamperes, 𝐼 sub B is equal to 8.148 times 10 to the negative five
multiplied by 1000 milliamperes. Rewriting this as a decimal and rounding to three decimal places, we get our final
answer to the question that the base current is equal to 0.081 milliamperes.
